Find the energy autocorrelation function $\phi^e_{ff}(\tau)$ and the energy spectral density $\Phi^e_{ff}(\nu)$ of the signal $f(t) = e^{-\gamma |t|},$ where $\gamma>0$ is a real constant.


So, the energy autocorrelation function from the definition is:

$$\phi^e_{ff}(\tau) = \int^\infty_{-\infty} f^*(t) f(t-\tau) \ dt = \int^\infty_{-\infty} \Big[ e^{-\gamma |t|} \Big]^* \cdot e^{-\gamma |t-\tau|} \ dt \tag{1}$$

So, how can I proceed to evaluate the integral? In particular what do I do about the conjugated term?

For spectral density, it is given by $\Phi^e_{ff}(\nu)=|F(\nu)|^2$ which is the Fourier transform of the signal squared. FT of $f$ in this case is:

$$ \mathcal{F} \Big\{ e^{-\gamma |t|} \Big\} = \int^\infty_{-\infty} e^{-\gamma |t|} e^{-j 2 \pi \nu t} \ dt \tag{2}$$

We also know that energy spectral density is the Fourier transform of the energy auto-correlation function. So if we evaluate (1) and take its its Fourier transform we should arrive at (2). Is that right?

Any help would be greatly appreciated.

  • $\begingroup$ i am assuming that $\gamma$ is real and positive. (oh, i don't need to.) you need to recognize that $$ e^{-|x|} = \begin{cases} e^{-x} \quad \text{for } x \ge 0 \\ e^{x} \quad \text{for } x \le 0 \end{cases} $$ and break your integral up accordingly. and that "breakup" works a little differently if $\tau$ is less than zero than if $\tau>0$. $\endgroup$ – robert bristow-johnson Jun 2 '16 at 1:17
  • $\begingroup$ Do you mean for evaluating the Fourier transform? The breakup would be$$\int^0_{-\infty} e^{j \gamma 2\pi \nu t^2} dt+ \int^\infty_{0} e^{-j \gamma 2\pi \nu t^2} dt.$$ But we can calculate the energy spectral density $\Phi^e_{ff}(\nu)$ of the signal using both (i) $\phi^e_{ff} (\tau),$ and (ii) directly from $f(t)$ (using Fourier transform). I am trying to find $\Phi^e_{ff}(\nu)$ using the first option by first finding the energy autocorrelation function. $\endgroup$ – Marc Jun 2 '16 at 2:17
  • $\begingroup$ Humph! Where did that $t^2$ in the exponent come from? $\endgroup$ – Dilip Sarwate Jun 2 '16 at 2:44
  • $\begingroup$ @DilipSarwate Oops, I meant to say$$\int^0_{-\infty} e^{(\gamma-j2 \pi \nu)t} \ dt+ \int^0_{-\infty} e^{-(\gamma+j2 \pi \nu)t} \ dt.$$I think if we then square this we end up with $\Phi^e_{ff},$ but my question is how do you find psd by first finding the energy autocorrelation $\phi^e_{ff}$? $\endgroup$ – Marc Jun 2 '16 at 4:34
  • $\begingroup$ leave the square out. all's i am saying is that you need to remove the absolute value "$| \cdot |$" from any of the integrals. and you can do that by considering where the $|x|$ function flips from being $-x$ to being $+x$. $\endgroup$ – robert bristow-johnson Jun 2 '16 at 4:46

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